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Did the X-15 ever conduct a point-to-point flight?
The North American X-15 rocketplane flew at up to Mach 7 and at very high altitudes, it was also capable of spaceflight but these were quite steep parabolic patterns. At Mach 7 it flew leveled and my question is whether the X-15 ever landed on another airport than the one it launched from (attached to its B-52 carrier plane so actually the airport where the B-52 launched from). Wikipedia seems to be silent on this.

The X-15 landed on one of many dry lake beds near Edwards AFB, it never landed back at Edwards.

@GdD It didn't land on a runway ever?

It landed on skids @Giovanni, it was designed to land on dry lakebeds, not on paved runways.

@GdD The SpaceShipOne and Two land on skids too, but on runways. :-)

Those are skids designed to land on a runway @Giovanni, and it's just the nose which has a skid. The skids on the X-15 were designed to land on lake beds.

@Giovanni: SSO and SST have nose skids but main wheels. Almost all of the weight of an airplane is on the mains, and the nose is kept up in the air quite a bit longer than the mains. The X-15 was the other way around: main skids and small nose wheels.

@JörgWMittag I see.

my question is whether the X-15 ever landed on another airport than the one it launched from

That depends on what you mean by "landed", "airport", and "another".
On its free flights, it never landed on another airport than the one it started from. However, it also never landed at the airport it started from. In fact, it didn't land at an airport at all.
It always landed on a dry lakebed near the airport. Whether you consider them part of the airport or not, is a matter of definition. As is the question whether you consider a dry lakebed an airport. I believe that the lakebeds were prepared with temporary runway markings, and they also may have cleared the "runway" from rocks and debris.
However, on aborted flights or captive carry flights, when it was never released from the carrier aircraft, it obviously landed attached to the carrier aircraft back at Edwards AFB.

Why didn't it land on runways? Were they simply too short for the X-15 to decelerate in time?

I don't know, to be honest. Some theories: It landed on skids, probably because of weight reasons. Skids and asphalt don't mix well. Also, a lakebed is easier to hit than a runway. The Space Shuttle, for example, also landed both on the runway at Edwards AFB as well as Rogers Dry Lake.

The X-15 on this photo has a gear, not a skid. Did that depend on the X-15's version? https://commons.wikimedia.org/wiki/File:X15A2_with_tanks.jpg

X-15A-2 was heavily damaged in a crash, and rebuilt with significant modifications, including the tanks and a lengthened fuselage. I can't find anything about changing the rear skids to wheels, though. Those look a bit flimsy, they might be transport wheels for moving it around on the ground.

According to Aerofax Datagraph 2 North American X-15 p.11 that is indeed a dolly used for moving the vehicle on the ground. Note that in the picture asked about in the comment, the lower half of the ventral fin is not installed. Good picture on page 52.

@JörgWMittag It's some sort of transport wheel. see here: https://commons.wikimedia.org/wiki/North_American_X-15#/media/File:X-15A2_2.jpg


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Are there jet aircraft for whose flights astrodynamics must be taken into account?
Civilian jet airliners don't need to push their yokes down to follow the curvature of the Earth because within the atmosphere the curvature is accounted for by automatical systems, right? I wonder whether there are jet aircraft (from the military, obviously) that actually have to push their yokes down for instance. The SR-71 Blackbird flew very fast (more than Mach 3) at very high altitudes (above 80,000 ft). Did Blackbird pilots have to be careful not to fly too much up due to the Earth's curvature, respectively not to cause a stall? Did Blackbird or other jet's pilots need to take other astrodynamics into account, such as rotational forces or other pitch concerns?

Comments are not for extended discussion; this conversation has been moved to chat.

It's the same question as for the ISS in orbit, gravity is pushing the yoke down for you. When you fly level you're actually keeping your distance from the earth center constant.

But the nose pitch remains the same relative to the stars while not remaining the same relative to the surface. Therefore, it goes more and more up, till the threat of a stall.

No you adapt your nose pitch to remain at constant density altitude.

The ISS is moving at orbital velocity - crazy fast. Even the fastest aircraft is moving, even with Earth's rotation, far slower. Apples to Oranges comparison here.

The SR-71 went to 3500km/h; low Earth orbit is at 28000km/h. Indeed apples to oranges.

He is asking if an airplane will pitch up if flying level, thinking that the airplane will go on a straight line and at some point leave earth atmosphere by doing so, which to me is similar to leaving earth gravitational field. I'm answering that even the ISS at 400km and orbital velocity around our planet is not capable of doing and then explain what flying level actually means...

@RalphJ The fastest jet aircraft was the Blackbird. The fastest rocketplane is the SpaceShipTwo which actually flies into space and doesn't adapt its nose pitch automatically up there anymore.

@MaximEck So there's enough density up there for the Blackbird to remain a leveled orientation automatically, even when flying at Mach 3 and above 80k ft?

For the back bird definitely, the space limite is actually defined as the limit at which air density is so low that an airplane will have to will at orbital velocity to generate enough lift to fly level. Below that airplanes fly normally. The spaceship burn all its fuel during ckim with a nose up attitude, then it coast using its velocity to counteract gravity during some time which will bring it back down at on point and then in fall pretty much like a feather at first and then start to fly once low enough. But they don't really control the trajectory in space as they don't fly.

@MaximEck Wouldn't a too high speed become difficult at a certain point, even deep within the atmosphere and with the necessary heat protection? Perhaps I should ask that as a separate question.

@Giovanni look, this is very simple. Gravity and air density turns the flight path down. Then, because of the plane's aerodynamics, it prefers to turn "into the wind", like a wind vane or shuttlecock, so it naturally pitches down.

@Abdullah Why didn't you say that immediately? And unfortunately I can't upvote comments but I would yours. Still, a part of the question remains, namely at what speed would navigation become not good, but that's in a different question anyway.

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Can military aircraft send out false ADS-B tracking information?
Can most military and VIP aircraft out of Andrews AFB transmit false ADS-B signals?
Here is the ADS-B tracking of Navy P-3 Orion that seems to be violating Iranian air space tonight if the ADS-B is correct. An Iranian fighter was launched heading in it's direction but did not approach it. Two nights ago I was watching on globe.adsbexchange and flightaware the tracking of AF1 was transmitting. It was about 2 miles out from the air field for a landing and the ATC tower noted to ground crew VIP traffic is 10 miles out. It showed it landing when in fact it was still in the air. I know they are allowed to turn off their ADS-B anytime they need to for security reasons.

Can they, I'm certain they can. Any ADS-B transmitter sent false info will send out false results, the real question is do they. As far as the accuracy of the track being reported, this is a good Q&A on the various problems online trackers have with ADS-B.

I get army drones on my ADS-B antenna, just FR24 will not display the tracks. ADS-B is good for anti-collision, so for training operation this is important (think about public reaction for a mid-air collision with a medair). OTOH for security, it is good not to publish them online.

Any ADSB signals that are going through an Iranian receiver on their way to FR24 or similar sites, are also subject to alterations by Iran.  In fact, given their record of duplicity, I'd treat every ADSB feed coming from Iran as unreliable and highly suspect at best, and a probable fabrication in case of anything weird or dubious.  (And a P-3 flying over not just Iranian airspace but also their landmass is fully in that category.  A B-2 over Iran, I'd believe it but you wouldn't see it; a P-3 over Iran... doubt!)

It is certainly possible, and also happens with some regularity. Here is an article discussing a recent USAF RC-135 flight over the South China Sea in which the RC-135 changed it's Mode S identifier to that of some Malaysian-registered plane.

At some point, the plane’s Mode-S number suddenly changed, from AE01CE to 750548. That’s the ICAO number for an unknown Malaysian aircraft. The RC-135W, call sign RAINY51, then flew a racetrack pattern between Hainan Island and the Paracel Islands, an archipelago in the South China Sea that China claims, but whose ownership is in dispute.

https://ph.news.yahoo.com/u-spy-plane-impersonates-malaysian-130000120.html
Relevant twitter thread linked in the article, with ADSB Exchange screenshots
https://twitter.com/steffanwatkins/status/1303776525083381760

Thanks for info. I went back and look at larger screen capture and noticed the P-8 was actually recorded as a MLAT signal source and not ADS-B.

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Will the combustion chamber I have made work as intended

I am trying to make a fairly simple electric jet engine, this is the kinda the design I came up with. I just want to know if this is a sound design for my combustion chamber. Pretty new with jet engines and sheet metal working. So I am not really sure of the things that could possibly go wrong.
More context, Everything is stainless steel and I am using butane gas, and the the fuel line sticking through is also stainless steel. No spark plug, will ignite the thing by a lighter and the compression is done using a electric motor driven impeller, not shown here.

With an electric motor for your compressor... do you have a turbine section at all?  What benefit do you foresee by using butane instead of jet fuel?  Perhaps it would be better to understand existing jet engines more thoroughly, before working to design one from scratch.  There is 80+ years of experience with designing, building, and operating turbine engines out there, and you probably don't have enough time to make all of the mistakes yourself by reinventing things that have gone before.

@RalphJ Ah its just to get my hands dirty, see some fire shoot out of the contraption. Learn a couple of things before I try harder projects

Since this is effectively a butane stove 'work as intended' is likely - the primary safety risk is the butane cylinder - local and remote shut off valves, and a check valve so as cylinder pressure falls turbine cannot pump hot air INTO the cylinder making things explosive. Would expect it to set things on fire rather effectively. If objective is to learn things rather than make a flame thrower would suggest pressure and temperature monitors and a custom controller - design goal is to control fuel and turbine speed to manage temperature just below point exhaust goes engine rich.

Above comment only applies to design show - if it gains a compression turbine or other outlet restrictions to the area where fuel and high pressure air are mixing it moves from 'big camping stove' to 'pipe bomb' quite quickly, classically occurring during startup https://en.wikipedia.org/wiki/Rocket_engine#Hard_starts

very insightful @GremlinWranger the type of answer I was looking for. Thank you so much, if you post that il accept it

On which side is the compressor and the turbine in your picture? Is that picture representing one can in a row or the whole chamber?


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Can an American citizen schooling abroad join the airforce?
I am a citizen of the U.S who is studying abroad from the age of 9.
If I am to Apply to become a test pilot will there be any impediments?
Due to security issues as it deals with classified aircrafts.
Aviation is something I really want to get into and I plan on taking aerospace for my college course, becoming a test pilot is an overlap of both worlds and is something I always wanted to be. I feel such reasons could be a hindrance.
Any links and answers will be much appreciated and If this question is not within guidelines a point in the right direction would awesome.

In which countries have you been studying? It probably won’t make any difference. Also, you do realize that there is no way to apply to be a test pilot ar entry into the military. You would have to serve as a pilot first before getting the opportunity to be a test pilot.

Why do you think a US citizen wouldn’t be allowed to join the US Air Force? Whether you have a shot at becoming a pilot (much less a test pilot) is a separate matter.

It is very possible for you to become a military test pilot. Follow these steps. Move back to the US. Get into a university. Join AFROTC. Graduate from both. Pass the FBI background check. Join the military. Become an officer or warrant officer (for helicopters). Become a military pilot. Serve and gain experience as a military pilot. Apply for test pilot school. By the time you apply to test pilot school, it won’t make any difference that you spent 10 years living abroad.

You may still be able to become a military test pilot without moving back to the states or graduating ROTC before your background check. It just makes the process more difficult. Unless, of course, you were overseas because your parents were serving a diplomatic or military mission. Living in an allied country or a community of American expats would help the process of the FBI background check as well. Read about https://en.wikipedia.org/wiki/Andy_Thomas

@Dean, indeed, that's a valid answer, you should have posted it as such.

In which countries have you been studying? It probably won’t make any difference. Also, you do realize that there is no way to apply to be a test pilot ar entry into the military. You would have to serve as a pilot first before getting the opportunity to be a test pilot.
It is very possible for you to become a military test pilot. Follow these steps. Move back to the US. Get into a university. Join AFROTC. Graduate from both. Pass the FBI background check. Join the military. Become an officer or warrant officer (for helicopters). Become a military pilot. Serve and gain experience as a military pilot. Apply for test pilot school. By the time you apply to test pilot school, it won’t make any difference that you spent 10 years living abroad.
You may still be able to become a military test pilot without moving back to the states or graduating ROTC before your background check. It just makes the process more difficult. Unless, of course, you were overseas because your parents were serving a diplomatic or military mission. Living in an allied country or a community of American expats would help the process of the FBI background check as well. Read about Andy Thomas, an astronaut from Australia that did not even become a US citizen until he was 30.

Some clarifications:  USAF doesn't have warrant officers (Army does). USAF helo pilots are commissioned officers. Study for BS in engineering or math. In AFROTC, apply for pilot school. Selection is before you graduate, but you can keep applying after graduation if you don't make it while in ROTC. Background check is done before graduation. When you graduate, you get commissioned the same day. At that point, you're in the military. He can also apply to the AF Academy, but getting a nomination while living overseas is a challenge. I suggest contacting an academy admissions counselor.


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I have an idea about how to resolve a conflict between the pilots and an automated flight control system. Does it make sense?
I am thinking about the 737-MAX crashes.  There is a similar question at Is it common for automated control systems to use non-redundant sensors?  There was an article in yesterday's Seattle Times In key step for potential late-November ungrounding of Boeing’s 737 MAX, FAA details minimum pilot training which references Why Boeing’s emergency directions may have failed to save 737 MAX.
My thinking is that if the pilots are applying a force to a control which is near the limit of the strength of a human being, then the control system ought to recognize that the pilot is fighting with it.  If the control system recognizes that the pilot is fighting with it, then it ought to assume that the pilot knows what he or she is doing, and let the pilot do what he or she wants to do.  There might be an indication along the lines of "Are you sure you know what you are doing?".
I order to recognize that the pilot with fighting with the system, I propose adding a strain gauge to the controls.  Normally, one should be able to fly the airplane with one hand.  If a pilot is using two hands to fly and is using large forces to do so, the strain gauge will detect that (technically, the pilot is stressing the control by applying a force, and the strain gauge is measuring the deformation of the parts the pilot is forcing), and that will tell the automatic system to "let go" and do what the pilot wants to do.
Does this sound like a good idea?  I am neither a pilot nor an aeronautical engineer.  However, I once worked as a computer programmer in a test lab and I wrote software to test strain guages.
Thank you

Hi! Don't get this wrong, but this isn't really a new idea. Autopilot disengage on strong control movement is already a thing in design. It has caused at least one accident, too (but has possibly prevented some).

Does this answer your question? Can the autopilot (or stall avoidance system?) of the B737 MAX 8 be overridden by sheer force?

Questions shouldn't be downvoted simply because they have been considered before.  It was well though out, and well worded.

As far as I know the B737 autopilot responds to light yoke inputs. If you push forward only slightly the autopilot remains engaged but autopilot trim orders for nose up trim are cut off. Similarly it works when you pull on the yoke slightly. (Or was it a different aircraft?) However this system can't be used because the B737Max is aerodynamically too unstable and MCAS was added to increase the column forces to make it feel more stable and to assist the pilot close to a stall. If the system would shut off when the pilot pulls up you can just not install the system and call it a day.

The question was downvoted because it does not accurately portray the problem with MCAS. MCAS is not an autopilot. It is an automated electric trim control. There were already procedures in place to deal with an electric trim that is not doing what you expect it to. You just turn it off. Both MCAS accidents would have been prevented if they just followed the established procedures that they should have been trained for.

@Mike Sowsun, This site has LOTS of questions that don't accurately portray a problem.  This one is not nearly as bad as some...  Part of the job of a knowledgeable expert is to clarify for the lay person any misunderstanding in the original question as part of the answer.

You actually bring up an interesting point.
The force gradients in the elevator system in a non FBW airplane are set by  some form of bungee system, like springs, that you are pulling against.  The hydraulics drive the control surfaces and there is no actual tactile feel, beyond what the feel force system imposes (think of fully hydraulic power steering in a big earth mover where there is no mechanical feedback to the steering wheel other than some form of spring centering unit).
The feel force is typically set to increase to about 50lbs at full travel, less at lower speeds and with flaps out, maybe 30lbs.  So the "two hand pull" required to get full elevator is there by design.  The force gradient is necessary to give the pilot a sense of how hard to pull and tries to approximate the forces on an airplane with mechanical controls.
You could easily design the feel force unit in the airplane to reduce the spring load to make the stick lighter for whatever purpose you want, and modern airplanes with the newer flight data recorders already have force sensors in the control linkage to provide data to the Flight Data Recorder, so the hardware is already there.
Problem is, the force gradient is needed to keep from overcontrolling, and either overstressing the airframe when above "maneuvering speed" or going into an accelerated stall when below it, from making large control inputs.
Where that kind of feature could be useful however, is during stab trim runaways, especially runaways to full nose down, and what happened in the MAX events, which was more or less similar to a ND stab runaway (the MAX's inadequate MCAS architecture got away with it by taking credit for the existing stab runaway procedure as a mitigation).
I've done those sorts of runaway scenarios in the simulator during recurrent training for the CRJs and the full nose down runaway leaves you wishing the feel force could be reduced as you have to continuously hold 20 or 30 or 40 pounds to keep the nose up.  It's very unpleasant (in the RJ, the fix was to get the flaps out one way or another, which drastically reduced the required feel force thanks to the pitching moment of the flaps reducing trim speed,  plus the feel force reduction that results from flap extension).
To implement what you suggest, you would have to have the feel force system set up to respond to an uncontrolled stab trim runaway by reducing the feel forces.  I'm not aware of anyone who has done this, and certification would probably be quite difficult (there will be many many unintended consequences that have to be ironed out, such as how to get the system to decide what a legitimate runaway is in a reliable way, and then how to get around the basic feel force requirements in a temporary way).  The certification hurdles are the likely reason this concept has never been done.

Failsafes within failsafes multiply modes of failure.  Consider the drone that, when it loses its GPS signal, retreats to a hover at 100 ft.  Guess what happens when it's powered up "safely indoors" in a shed with a steel roof.

Off topic, but I once did the opposite to save a radio-control sailplane whose elevator servo had failed (diagnosed after landing).  Fiddling with the flaps regained just enough pitch control to land smoothly.


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VTOL Propeller Sizing and Executing a transition from VTOL to Forward Flight mission profile
Hi I am an engineering student, currently in the process of designing a compound UAV drone capable of both vertical and forward flight. I've made significant progress, having completed initial gross weight estimation, Wing Loading and powerloading estimation, wing, fuselage, tail, powerplant, and landing gear design, as well as weight estimations. Moreover, I've solved for the center of gravity at each component location using equations from Daniel Raymer and Mohammad Sadraey, resulting in minimal CG travel.
This is the UAV that I have designed:

However, all these calculations have been based on single-engine estimations from Raymer as advised by my professor. My current hurdle lies in the VTOL (Vertical Takeoff and Landing) mission profile. I'm unsure about the rule of thumb for determining the diameter of the VTOL propellers and whether it's a critical aspect. Unfortunately, my professor hasn't provided much guidance regarding the VTOL motors.
Another challenge I'm facing is how to execute the required mission profile, as depicted below:

According to my professor, I need to gradually reduce the RPM of the VTOL mechanism while increasing the RPM of the pusher propeller. However, I'm concerned that this approach might impose excessive torque on my VTOL propeller. My alternative proposal is to go power-off once a certain height is reached and then increase the RPM of my pusher propeller as the aircraft glides. Nonetheless, I'm uncertain if the airfield space would be adequate for this specific mission profile.
Any advice or guidance on these issues would be greatly appreciated.

I'm unsure about the rule of thumb for determining the diameter of the VTOL

The rule of thumb is termed momentum theory, by which the thrust $T$ generated by a rotor is:
$ T = k\sqrt[3]{2 \rho A P^2} $
where $A$ is the rotor area, $\rho$ is the density and $P$ the power needed to spin the rotor. This equation is strictly valid only at hover but being the highest thrust normally required in hover, then it's just perfect for your purpose. The $k$ is a coefficient taking into account the limitations of the momentum theory:  a value of $0.5$ can be used in a conservative way.
Obviously $T$ equals weight $ T = W = mg $, where $m$ is the mass and $g$ the gravity. Just divide it by the number of propellers you have in mind (4?).

According to my professor, I need to gradually reduce the RPM of the VTOL mechanism while increasing the RPM of the pusher propeller. My alternative proposal is to go power-off once a certain height is reached and then increase the RPM of my pusher propeller as the aircraft glides.

Just simulate both and compare the results, it might be that your approach work as well.

Theres lots of equations, here's one with empirical corrections. These calculations can be fairly inaccurate. You can always find the motor/prop combo that provides enough thrust for the weight. For 4 props at 50% throttle.

Link for equation
For transition you need to work out how much lift you gain relative to velocity squared. The thrust curve vs. RPM isn't linear either :)

I cannot upvote the answer related to blade momentum theory due to my low reputation ;) but it is way to go. Moreover, the following paper may help you with your project: https://www.mdpi.com/2226-4310/11/3/200
VTOL may need forward velocity to glide and transition is required to be executed gradually so that flow can be established over the wings. In the absence of airflow, a flying vehicle may fall into a flat spin.
Obviously, one can simulate both cases but that will require some CFD data at high AOA as linear methods lack the complexity of a separated flow over a falling aircraft.


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How do aircraft land and take off on an aircraft carrier?
The average aircraft carrier length is 1000 feet, and the runway for a really small airport is around a mile long. How do they take off and land?

Welcome to aviation.SE! See this question for landing, and this one for taking off. Does that help?

Even the most basic research would tell you that takeoff uses catapults, landing uses tailhooks & arresting gear.  Don't think you know much about small airports, either :-)

A small airport could have a runway or runways with lengths between 3000 and 12,000 feet. What you probably mean is a short takeoff and landing runway. In the US civilian world, a short takeoff and landing surface would be anything under 3000 feet for most people. But, it could be as short as 1200 feet. It really is defined by the aircraft used. I regularly takeoff and land on a runway surface of a little over 3000 feet using normal (not short) landing procedures. There are some private airfields with runways under 1000 feet.

An awareness of the existence of aircraft carriers, generally, comes packaged with enough supplemental basic information to answer the root question.  Catapults and arresting cables are the short answer.  Plenty of information out there, downvoting for lack of effort.

Welcome to AviationStackExchange. While this is a forum that welcomes the inquisitions of all aviation enthusiasts, there is an expectation that some level of research was done on the subject by the question’s poster. You may find it useful to add to your questions the lines of research you have done so far, and even your level of understanding of the subject. Then, ask your question as clearly and concisely as possible to includes the specific areas of your research on which you are confused. Otherwise, overly broad questions of readily accessible information may be deleted.

the planes are basically thrown down the runway with a steam-powered catapult that boosts them up to flying speed by the time they reach its end. For landing, they use a tail hook on the plane that catches a cable or series of cables stretched across the runway; the cable(s) then bring the plane to a quick halt.
The stresses imposed on the airframe by all this violent action are great, requiring very strong construction. The hazards presented by carrier operations are great too; a malfunction of the catapult which causes the plane to not reach flying speed will throw the plane off the bow right into the path of the ship, which then overruns the plane. And if the tail hook misses all the cables, the plane is at risk of running off the runway and falling into the water.

There are multiple different systems in place. The most widely-used ones are:

CATOBAR (catapult-assisted takeoff but arrested landing): the airplane spools up its engine(s) to full thrust while being held in place by a restraint system, then gets literally catapulted off the deck with a sled that is attached to the nose wheel gear and "slings" the airplane forward and accelerates it to takeoff speed. For landing, the airplane extends a tail hook that catches a steel wire on the deck which "pulls" the airplane to a stop.

STOL (short takeoff and landing): some airplanes are capable of normal takeoff and landing on the runway lengths provided by an aircraft carrier. This is mostly no longer in use, but it was the way things worked in WW I and WW II. The world records for STOL airplanes at the moment are about 10 feet for takeoff and under 10 feet for landing.

VTOL (vertical takeoff and landing):  some airplanes are able to take off and land vertically. Some examples are the McDonnell Douglas AV-8B Harrier II "Jump Jet" and the F-35B variant of the Lockheed Martin F-35 Lightning II.

STOVL (short takeoff / vertical landing): taking off vertically uses a lot of fuel and limits the weight of the airplane (meaning they cannot take as many weapons with them), therefore it is more efficient to have a short takeoff roll and take off with at least some forward speed.

Ski Jump: at the end of the deck, the very end of the runway is curved upward, this gives the airplane additional upward momentum and accelerates it upwards into the air. Typically used for STOVL airplanes. The most well-known examples are the British aircraft carriers of the last decades.

In all of these different modes, the aircraft carrier can always do two things that an airport cannot do:

The carrier turns directly into the wind (or in case of an angled flight deck at such an angle that the runway points directly into the wind). That means, an airplane never has to worry about taking off with a crosswind or tailwind, it will always have a headwind, which increases the airspeed.

The carrier accelerates to its highest sustainable speed. So, even if there is no wind, the airplane already has some forward velocity relative to the airflow.


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Anyone knows where i could find the McMaster, H. C., “FAR 23 LOADS”, Aeroscience Software, 2.02 Edition?
I am currently conducting research on wing lift distributions and came across the existence of a software tool called "FAR 23 LOADS," developed by McMaster, H. C. (Aeroscience Software, 2.02 Edition). This software seems to be specifically designed for analyzing  loads based on FAR 23 regulations.
However, despite my efforts, I have been unable to locate any sources providing access to this program. So far, I have only come across its user manual. I am genuinely intrigued by the capabilities of this software and would greatly appreciate any guidance on where or how I could gain access to it.
If any of you have information regarding the availability or any alternative software tools that can aid in analyzing wing lift distributions, I would be immensely grateful for your insights.
Thank you all in advance for your time and assistance.
Best regards,

You can request a quote here. For the history of the IP re-purchasing from McMaster, see here.


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TEB-NAS Midsize Jet max weight
We have hours with Sentient Jet and typically fly their Midsize jets -- Hawker 800, Citation XLS, Lear 60 etc. Usually only the XLS will work for us due to large luggage.
Can someone help with me with maximum weight for our trip combined with our routing?
I am not sure we will fit or be able to do non-stop due to pass/weight.
Last time we flew 6 pass + little less luggage and had to make a stop half way in South Carolina. I am thinking at least we will need to do this again.
As an alternative - I might call Sentient and ask if they can source us a Citation Latitude or Legacy 450 which I think will work? But they may consider these Super-Mids?
Otherwise we can upgrade to the Super Mid size category which of course is quite a price jump.
See info below and let me know any thoughts. Thank you....

Route: TEB to NAS (Teterboro into Nassau Bahamas -- Just about 1,000 nautical miles?)
Passengers: 7 adults + 1 child (about 1,300 lbs) + 2 pilots (400lbs)
Luggage: 8 mid/large bags + 3 golf bags (about 550 lbs)
Total weight excluding fuel = 2,250 lbs


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Unusual pattern over English Channel
I have just spotted this Beechcraft flying an unusual pattern over the English Channel (late at night). It has no call sign etc. I'm thinking maybe a military aircraft looking for illegal migrant boats with thermal imaging (altitude is a constant 1900ft). Any thoughts?

This is a duplicate of this question about the same aircraft SRD244B

Note that the flight does not have "no callsign;" rather, the specific website/app you were using voluntarily chooses to not display the callsign. From the ADS-B Exchange replay, we can see that the aircraft registration is  G-HMGA and it was using the callsign SRD244B. "SRD" is the identifier for HM Coastguard and on the radio the callsign would sound like "COASTGUARD TWO TWO FOUR BRAVO."

According to Wikipedia, HM Coastguard does not perform the functions of "Customs/Border Control, Fisheries Control and Law Enforcement." This leads me to believe that the aircraft would not have been searching for migrant boats simply on general customs/border control principles; they could have been searching for a boat in distress, as "Search and Rescue" is one of the Coastguard's functions. Or it could be that Wikipedia is not correct, or it could be that the Coastguard does sometimes support other agencies' missions.
Of some passing interest, the registered owners of G-HMGA are not the Coastguard themselves but a private company called "ACH London Ltd." This seems to be a holding company associated with 2Excel Aviation. Apparently the Coastguard does not directly own and presumably does not directly maintain this aircraft, even though it is painted in Coastguard livery.

Lloyd Bell at JetPhotos

2excel all but admit they do "law enforcement" and "surveillance" (read as: search for migrant boats) using these exact aircraft: https://www.2excelaviation.com/2excel-aviation-delivers-life-saving-surveillance-capability/ . In addition the area they were operating in is almost constantly NOTAM'd for this exact purpose.

Could be any number of reasons. Polling ocean buoys, aircrew training, ship crew training, google map something or other.
Or, searching for a maritime craft, doing something it should not be doing in the English CHannel.
It took off from Humberside, and is presumably returning there.

RAF commander who doesn't know that Operation Fortitude is over.

HM Coastguard aircraft/flight. Wikipedia says that Customs/Border Control, Fisheries Control, and Law Enforcement are the three functions not carried out by the Coastguard, but I don't know if that means they never support the agencies who are responsible for those functions.


text:
Can jet engines stay running without electrical power?
A piston-engine aircraft will continue to run as long as its magnetos still work, even if there is zero electricity in the battery and the alternator is broken, because magnetos are independent of the aircraft's electrical system. Since turbine engines don't have magnetos, do jet aircraft (including turboprops) work the same way?

I think a clarification of the question could help narrow down whether the question is asking about any electrical power at all (i.e. purely mechanical), or if the question is only about electrical power provided from a source external to the engine housing itself.  The questions so far seem to indicate that electrical power is typically required, but is usually provided internally by the engine itself.

Yes, a jet engine can operate without electrical power.  Or at least they used to be able to...
A jet engine uses igniters (like electric spark plugs) to initiate combustion in the combustor.  Once initiated, the igniters are not needed anymore and the combustor should stay operating as long as the engine is operating.
Modern jet engines use a FADEC (Full Authority Digital Electronic Control) to control the engine.  The FADEC is a computer that monitors everything about a jet engine and simplifies its operation to the pilot.
Older jet engines used different combinations of mechanical, pneumatic, and hydraulic control.  A modern FADEC is better all around, improving engine performance, life, and safety.
Modern FADEC are redundant and are extremely reliable.
The FADEC requires electricity, but it will have a separate generator analogous to the magnetos.

Are the fuel pumps mechanical or electrically operated? If they're electrical, the combustors will stay lit while they've got fuel, but as soon as the pumps stop, the fire will go out. No?

@FreeMan There's gravity feed from wing tanks for underwing engines.

@FreeMan The high-pressure fuel pumps within the engine are mechanically connected to the accessory gearbox and will continue to run. The low-pressure fuel pumps in the tanks might not, but as Therac said, gravity and suction from the high-pressure pump is enough for under-wing-mounted engines. I think tail-mounted engines might have problems though.

From https://www.faa.gov/sites/faa.gov/files/2022-01/Full%20Authority%20Digital%20Engine%20Control%20(FADEC).pdf If the FADEC fails, the engine fails. However, system redundancy makes it much less likely for a FADEC system to fail than a traditional magneto system. In fact, a double magneto failure is statistically more likely than a FADEC failure.

@user253751 I think the context of modern engines using a FADEC (which uses electricity) makes it clear the possible exception I'm referring to.  The OP allows a spark ignition piston engine to operate 'without electricity' because it has a self-contained power source (magnetos) independent of the aircraft's electrical system.  If the same allowance is made for a FADEC's independent power source, then Yes, it can operate independent of the aircraft's electrical system.  If however, all electricity is forbidden from use in this question, then no a modern jet engine will not operate.

Jet engines have magnetos too (alternators with permanent magnets). Combustion can be sustained without igniters, but a permanent electrical power supply is required to feed engine electronic components, in particular each engine controller (EEC, ECU, FADEC).
This controller adjusts the engine parameters, according to the thrust demand selected by the crew with the throttle (fuel flow, variable vanes...)
Each engine has its own dedicated electrical alternator for the controller. If this power supply is not operative on the ground, the crew cannot take off (see A320 minimum equipment list).

(Source: CFM56-7B Engine systems)
A controller has two redundant channels fed by different sets of stator windings in the alternator. The controller is initially fed from the aircraft power supply, and the dedicated alternator takes over when the engine spins (> 12% N2 on CFM56).
If any set of windings fails:

The corresponding controller channel is switched back to the main aircraft supply (IDG, or whatever currently feeds the main electrical bus). If this happens while in flight, the flight can continue.

An ECAM message ENG 1(2) FADEC ALTERNATOR is displayed. If this message appears before takeoff, the crew is not allowed to take off.

Both the regular high-power generator (IDG) and the small controller generator are driven by the engine using the accessory gearbox, e.g for CFM56-7B of a Boeing 737 NG:

(Source: CFM56-7B engine manual)
This gearbox is located at the bottom or (like here) on the side of the fan case:

Source.
For more details about the accessory gearbox, see this question: Where is the generator in a large turbofan of a commercial airliner?


text:
Seeking Aviation Project Requirement Documents
I am involved in an aviation project and am in need of detailed examples of project documentation to guide our development process. Ideally, I'm looking for full project document sets that include System Requirement Documents (SRDs), Hardware Requirement Documents (HWRD), Software Requirement Documents (SWRD), Hardware Design Descriptions (HWDD), and Software Design Descriptions (SWDD). However, if full sets are not available, examples of SRDs alone would also be extremely helpful. Could anyone recommend resources, websites, or databases where I might find such comprehensive examples or templates, or even just SRDs? Any pointers to documentation or case studies in the aviation sector would be particularly helpful. Thank you!


text:
IRS error impact during different flight phases
I am working on my thesis on the Inertial Reference System and its errors and accuracy. I am struggling to find if the errors (biases, drifts, ARW, scale factors, etc.) have different impacts on the system during different flight phases (like if the position errors produced during takeoff could be bigger than during cruising). I need it with a source - a book, science journal, article, or something like that. Any help would be much appreciated.

There is a study which presents the errors vs. the maneuver parameters, by Paul D Groves for IEEE: Navigation using Inertial Sensors


text:
Lightoff to speak about an engine start
Why is the term lightoff (or light-off) used to describe the moment an engine ignite/is started ? I'm wondering why the suffix 'off' is used in an 'ON' sense (start, ignition) ? Is there a etymologycal reason, or an historical reason/origin for this use ?
I found this word specifically in aeronautical applications.

Just a follow up question... could it not refer to an annunciator light that extinguishes when the engine starts. I can only think of Oil Pressure warning, but I'm sure other lights go  to an "off" state?

@GregWoods  The aircraft I've flown all indicate the ignition with a rise in a temperature, either Turbine Inlet or Exhaust Gas temperature.  Other lights may extinguish either before or after ignition as the engine spins up, but the fact of ignition is observed via the swing in the gauge, not a light per se.  I think the top comment above is more to the point. (And I agree it could become a good answer.)

Note: there are a stack exchange about English (etymology, etc.). I think the question may be more relevant there. The links are on the top right icon (on the banner)

It's [english.se], @GiacomoCatenazzi (The shortcut for that is to type [english.se] and it's expanded into the full site name as a link.)

@GiacomoCatenazzi I've also posted my question on this SE, see link

@GregWoods It mights be an explanation, but not exactly in my application. The reason why I'm looking for the origin of the word.

The opposite word might be 'flame-out', to indicate engine shutdown.

It's one of those quirks of the English language. From the Cambridge dictionary - use of "off" to start a process:
"To begin by doing something, or to make something begin by doing something:

She started off the meeting with the monthly sales report.

I'd like to start off by thanking you all for coming today."

Another example: To make a firecracker explode, you "set it off".
So, literally,  it's the commencement of engine operation by the lighting of the flame in the burner can.

This would be best answered on English Language and Usage, but I see they already suggested to use Aviation. So this answer.
Off has several meanings, including distance or suddenness and being out. But as a noun it means the beginning. From the off means from the beginning.
Some phrasal verbs also use off in the sense start/launch/trigger/commence, including to light off:

To light off: To set fire to (fireworks or the like) in order to cause an explosion, usually for the purpose of entertainment. The kids are lighting off firecrackers in honor of Chinese New Year. Source.

The light off is the instant the engine is ignited. You'll find the same intent in other phrasal verbs like take off, lift off, start off, make off, set off and the derived nouns. Not forgetting the usual kickoff meeting every project starts with, which has a sense similar to light off.

I'm wondering why the suffix 'off' is used in an 'ON' sense (start, ignition)?

Note the different uses of off:

To light + off: The phrasal verb to light off, meaning to set fire to something, to ignite something.

To switch + off: Change the state of something to off.

As an example of off used to define some point in time: Liftoff is a term used for a spacecraft takeoff. For the US Shuttle liftoff meant the vehicle top reaching the top of the launch tower. Before this point the count down was negative, after it was positive: T-10s, liftoff, T+3s, etc.

Thanks for your answer. As a non native english speaker, I generally understand "off" as not working. In your examples take off, lift off, move off... I understand it as "away from", so it's comprehensible. But it's hard for me to realize when off means ON.

@Vincent: As explained in Wikipedia, a phrasal verb is idiomatic, the meaning cannot be inferred from the two words. Think about back up, which has different idiomatic meanings of which: move in reverse for a car and block for a drain. Other examples look after (take care), turn up (appear), tell off (reprimand). If you happen to be a French speaker, passer outre, or compter sur are constructions similar to English/German phrasal verbs.


text:
Calculation of density altitude
When using an e6b flight computer which temperature will I use with the pressure altitude to get the density altitude

What are the choices?

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer.

Use the outside air temperature at the location you are calculating density altitude for.


text:
What's the more accepted notation used to label sensors present in turbo-jet engines?
I'm interested in learning what the proper (more accepted) notation used to label sensors present in turbo-jet engines is.
Came across a few examples such the ones shown below but would like to learn if there is a 'standardized' methodology.

Temperature and pressure notation
Speed Notation

Sensors are identified in your picture by either P or T, for pressure or temperature, and the engineering station number.
Station numbering is standardized in SAE ARP 755: SAE Aerospace Recommended Practice 755 - Turbine Engine Performance Station Numbering. As you find yourself this standard has been extended to SAE AS 755 - Aircraft Propulsion System Performance Station Designation to take into account more complex engines
The main stations for a turbojet are:

Additional stations are used, e.g. for a turbofan. :

Source.
These numbers are used, for example, in this temperature-entropy diagram:

Source.
You can see the raise of temperature by the combustor between {3} and {4}. See this answer for explanations.

ARP755 has been superseded by AS755F [https://www.sae.org/standards/content/as755f/ ]

@user67034: Good find, I edited my answer to include this extension.


text:
Does ARINC-664 (AFDX) supersedes both ARINC-429 and ARINC-629?
Wondering if ARINC-664 (and ARINC-629) have backwards combability with ARINC-429. I realize -664 uses Ethernet while -429 is implemented using a two-wire data bus. However, wanted to see if any of the 32-bit ARINC-429 are "ported" upwards.

The network definitions of A429, A629, and A664 are independent of each other. If you look at the OSI model, one significant difference is that A429 includes the two translation/presentation layer options. One of which is the traditional  A429 messaging and the other a block data transfer. For the A429 messaging, it also includes the standard data dictionary. A629 and A664 are more traditional data bus standards for a full-duplex interconnection. A629 is a scheduled bus network. A664 is similar to Ethernet with topology and scheduling constraints to  make it more deterministic (guaranteed delivery and maximum latency). But A629 and A664 don't specify the data that's sent across the network. Since A629 is pretty much limited to legacy B777, I'll discuss just A664 as the multipoint network going forward, though the concepts are identical. (You probably wouldn't ever see A629 and A664 on the same aircraft. A664 has effectively replaced A629 as the high bandwidth bus.)
In the real world of an aircraft "network" it is common to use use two or more network types and have data move across them all. An aircraft has many sensors that provide everything from a small data stream of a few hundred bits/sec to video streams of MBits/sec. For cost and power/weight reasons, smaller low cost busses provide low rate data to data concentrators that forward the data onto the high speed/high bandwidth bus. For example, B787 uses CAN Bus, A429, and A664.
Following the above example, The overhead control panel is connected via CAN Bus to a gateway that connects to the A664 network. The status of each of the switches on the panel is a single bit in the CAN message which gets lumped into a big data word on the A664 network. Similarly, many legacy LRUs such as the Radio Altimeter just have a couple A429 output busses with a handful of data words. The A429 busses are connected to a data concentrator that packages the A429 labels into a larger block word and forwards it across the A664 network. Data can also go the other way. For high bandwidth devices such as the weather radar, it has its own A664 interface which can easily handle the bandwidth. In the A380, the A664 network is primarily associated with the display system which has relatively high bandwidth needs.
For convenience and interchangeability most data that has to go over A429 retains the standard A429 message formats. Within the A664 network, they are just opaque data blocks. For everything else, the aircraft integrator and the suppliers all have to agree on a data structure for the aircraft. It's a lot more up front work, but it allows for changes to be incorporated via software updates where the older A429-only systems required wiring changes which added a lot to the cost of change.


text:
Air density factor when calculating Airspeed
I am making my own attitude indicator / AHRS, like a Garmin G5. I am using two pressure sensors to measure altitude & airspeed via  static /pitot lines.
Now I know the formula for calculating airspeed is:   SQRT( (2 * ( Pitot - Static)) / density ) and many sources I find use 1.2 for the density factor, which should actually be 1.225 from my research.
But that density value is for air at sea level.  I'm thinking I should adjust the density factor based on altitude. For instance, at 10k feet, the density drops to about 0.91 kg / m3. Not a big difference but enough to possibly be a factor.
Anyone have experience doing what I am making?

If IAS is what you're looking for, you should simply not consider the reduction in air density due to altitude. As altitude increases and density decreases, the ASI will show a lower dynamic pressure (IAS) for the same true-airspeed, but the wings and rest of the aircraft will also experience the same reduced dynamic pressure. So use the sea level density irrespective of the altitude - if IAS (dynamic pressure) is what you want your ASI to read. But if you want TAS, you must account for density: $TAS = IAS / √σ$ (where $σ$ is relative density).

You need to look up and understand the differences between:
True air speed (TAS),
Indicated air speed (IAS),
Equivalent air speed (EAS), and
Calibrated air speed (CAS).

As someone else pointed out in the comments, the indicated airspeed should not be corrected for density. That's because its main functions is to tell you about the dynamic pressure, which is what really affects the aerodynamics and performance of the aircraft (yes, it's annoying for navigation, but the differences aren't great at low altitudes and besides, most people use GPS for nav these days instead of DR ;-) )
Airspeed indicators normally use a slightly different formula than the Bernoulli one you gave. It's:
$v_{IAS} = a_0 * \sqrt{5\left[\left(\frac{q}{p_0}+1\right)^{\frac{2}{7}} - 1\right]} $
Where $a_0$ is the ISA (international standard atmosphere) sea level speed of sound (340 m/s), $q$ the dynamic pressure (aka impact pressure) and $p_0$ is the ISA sea level pressure (101325 Pa).
Note that the magic numbers 5 and 2/7 arise because of the value of $\gamma$ (gamma), the ratio of specific heats for air, which equals 1.4.
It actually works out very close to Bernoulli for low Mach numbers. However, this way it can more easily be corrected in a standard way for compressibility effects (equivalent airspeed) and density effects (true airspeed) over a wider speed range.
There's more information, including some example code, see one of my other answers
You probably already know this but the numbers you are quoting (1.225, etc.) come from the International Standard Atmosphere - that's worth looking up for more information.
In terms of hardware, here are some examples from other people trying to do something similar to you: Experimental Avionics and here including source code.
This amazing guy made his own set of avionics from scratch for his motor glider.


text:
Why do planes fly at a high angle of attack when flying slow?
I've noticed that when a plane does slow flight, the pilot increases the angle of attack (AOA). What is the purpose of this?

F-18 - Slow Pass

Looks like a Concorde without the pivoting nose. See also Wikipedia about high AoA maneuvers.

@mins I'm pretty sure that's a military fighter aircraft, not any sort of airliner.

@nick012000: You might be right, there is no airline logo on the tail.

You should be aware that the picture you show (where I assume the plane is actually moving approximately horizontal)  is very much an extreme example. Most planes can't do that. While most planes do increase the angle of attack when they fly slow, it's by much less than this picture would indicate.

@mins That too, but look at the proportions of the aircraft. The cockpit, wings, and tail are all far too large for an airliner, proportionally speaking.

@nick012000: Now you say it I also realize there is only one pilot and no windows in the cabin.

@nick012000 and everyone... that's an F/A-18 Hornet. A fighter aircraft with an extreme flight envelope. This maneuver is likely part of an airshow or something similar. You can tell it's a Hornet and not the Super Hornet by the shape of the air intake, being straight up/down from this perspective, Super Hornet's intake would appear slanted forward 45* from the picture's perspective. So it's likely a F/A-18C Hornet.

Next time you're in a car, hold your hand out the window and make your hand flat like a wing.  Now tip your hand to change the angle of attack - notice how the force from the wind changes.

The lift a wing generates is directly proportional to

the density of the air
the square of the airspeed
the Angle of Attack (AOA) (more or less linearly) (until it stalls)

In non-accelerated flight, where the lift is roughly equal to the weight of the plane, the product V2AOA will be a constant, or AOA will be proportional to 1/V2.
As the plane flies slower and slower, the AOA has to increase to maintain the lift. Keep below the AOA for stall, result happiness. Increase the AOA beyond the stall, you have a bad day, if you don't have the height to recover.

Lift is (among other things) a function of the airspeed and the angle of attack of the wing. Hence, if you reduce your speed, you have to compensate the associated lift loss by increasing the angle of attack...

Is it just the AOA that is important or is the upward angle of thrust vector important too? In other words, would it matter if the engines where still pointing horizontally (and still behind the center of mass somehow)?

@johnDanger The component of the engine thrust which is pointed upward does contribute but this should be considered trying to make the best of a bad situation The wing is much more efficient and any thrust you divert from the horizontal to the vertical removes thrust that could otherwise be used to propel the much more efficient wing. You are supposed to "fly with on the wing". Ideally you have the engines continue to point horizontally while the wings increase in AOA but there are great mechanical challenges with a hinged wing or engine compared to just tilting the whole aircraft.

The vertical component of the thrust is important, and adds to lift... For angles of attack beyond the critical stall value, it may be higher than wing lift...

@DKNguyen But, for a small angle, the gain of vertical thrust is first-order while the loss of horizontal thrust is second-order (much smaller), so it seems it is always worth tilting the engine somewhat even though the wing is much more efficient.

@DKNguyen Considering that this is a military fighter aircraft, I'm pretty sure it's got enough thrust that it could take off on pure thrust like a rocket if they wanted to.

@nick012000 I'm talking about airplanes in general. And even if it's got the thrust, it doesn't mean it is the best way to fly. Fuel doesn't come from nowhere. You could walk around on all fours if you wanted to, but are you really going to unless you have no other choice?

@nick012000 The open question is what is the Lift/Drag ratio for any given aircraft.  If you know that number, you can do a lot to figure out whether it is the AOA or the vectoring of the thrust that gives the most effect.  It's really something you can't figure out in general.  You have to figure it out for the model of the aircraft.  I'd point out that non-military aircraft can fly slow too.  And they all rely on high angles of attack to do it.

In order to maintain level flight, an airplane must generate lift equivalent to its weight. The lift generated by a wing diminishes rapidly as the airspeed decreases. To compensate, the angle of attack must be increased to maintain the necessary lift. This can be done until the the critical angle of attack is reached, at which point things get exciting.

Where "exciting" = "stall" --> "fall from sky".

It's a great step that you know about the acronym AOA while still not understanding well how it works! That's nice; but anyway, try to exercise the answers here using your hand as a wing through the wind when someone (ELSE, lol) drives for you. You'll notice that in order to keep you hand flying at slower speeds, you'll have to increase the angle. This is the AOA that you mentioned.
For an aircraft it is the same.
If your mind is more mathematical, imagine flight as X+1=3; if you remove X from the equation, you must add more value anywhere so the wrong sentence 1=3 becomes the right sentence 3=3.
Remember that a higher AOA brings more drag, which in turn will require some more thrust to avoid the speed to decrease. Speed decreasing with the same AOA means that altitude is decreasing too.
And this bring us to another nice curiosity: airplanes can fly with TWO different speeds using the SAME amount of thrust. One, lower, speed is the one where the thrust matches the total aircraft drag when the bigger player is the angled shape of the aircraft facing the wind at a higher AOA. (It is, by the way, a flight that is very unstable; any input adding AOA will take a toll on speed and make you lose height; any lost height will give you back some better angle shape and increase your speed.) And the second, higher speed, is when the same thrust now matches the aircraft flying at a much better angle with a smaller 'shape against the wind', but the major player of the drag is now from the wings (lift-induced) and overall pressure against the flying body fighting a faster wind force.
Complicated, but always fascinating.

(Thank you the corrections, admin! Essential for a non-natural english speaker and great opportunity to learn more.)

As explained in various posts on StackExchange, at the end of the day the one thing that matters for lift is that air gets persuaded to move downwards (say, with velocity $v_\text{down}$). The downwards-acceleration of an air pocket of mass $m$ as it passes the wing corresponds to a momentum change of $m\cdot v_\text{down}$, and the same and opposite momentum is imparted on the aircraft, lifting it up.
If you decrease the airspeed, this is affected in two ways: the downwards component will scale down as well, and less air passes the wing per time unit. Both can be counteracted by increasing the AoA:

The steeper angle means the air is sent on a steeper down-trajectory, i.e. with a relatively higher downwards component.
The higher frontal area causes more air to be deflected downwards.

This assumes that the air keeps streaming along the wing smoothly. If you let the speed get too low and/or increase the AoA too much, the flow will stall, which decreases lift again while drastically increasing drag. Most planes can generally not cope with this well, but fighters like the F-18 can simply power through it. Note that the high AoA also makes the engine point downwards, so you get some lift even without relying on the wings at all.

In order to maintain level flight, the aircraft must push air downwards to create a force that supports the aircraft in the air. Lets consider a period of one second of time:

An aircraft going fast slices through a large amount of air, and only needs to push that air down a bit.

An aircraft going slow slices through a small amount of air, and needs to push this air down a lot.

Considering the wings as a wedge, the change in the amount the air needs to be pushed down can be achieved by varying the steepness of the wedge, this steepness is essentially the angle of attack.
This is a crude simplification, but gets you going in the right direction.

The four forces of flight are lift, thrust, drag, and weight. I don’t see pushing the air down on that list.

Smartypants  Explain lift then.

Oh, wait, you don't need to. Leftaroundabout already did that in his answer. Lift comes form, wait for it... air being forced down @JScarry


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What is that yellow square to the right of the main entry door of this Lufthansa A350?
My question is: is it just part of the paint job, or does this yellow square serve a purpose? I haven't seen this square on any other airlines.

It's a yellow square that says Willkommen / Welcome. In one case I have seen that additional text was written into the square with a black (removable?) marker.

Source
This website explains it like this:

A yellow welcome panel painted beside each of the passenger boarding
doors is now visible on each aircraft. The mark is a flat yellow
rectangle, with the word 'welcome' in both German and English,
accompanied by the iconic logo in the corner. What I love so much
about this - is that that bright yellow warmth is presented to every
passenger, right as they board every aircraft. Upon boarding, each
passenger is welcomed, setting the stage for the onboard experience
and hospitality that will follow. The crew can write personalised
messages to give a more human welcome to this otherwise mechanical
setting.

"bright yellow warmth is presented to every passenger" - yeah right. That's marketing for you...

"Bright yellow warmth" doesn't project much to me, but I tell ya' what, service & comfort on my transatlantic flight on Lufthansa to Germany was much better than the return flight service on a United aircraft. Even though they're both part of the same alliance.

@user253751 The door may have been designed by the Sirius Cybernetics Corporation. Share and Enjoy!

So it's a yellow whiteboard...

@jcaron: A yellowboard!


text:
How do rolling airframe missiles steer?
There are a certain type of missiles that roll constantly in flight, so-called rolling airframe missiles. Missiles found in the category range from various MANPADs, to the creatively named RIM-116 Rolling Airframe Missile. Wikipedia explains that by rotating in flight, only one pair of control surfaces are needed. How are the control surfaces directed in order to steer the missile, while constantly rotating?

It's actually simpler, you simply ignore azimuth error totally and only steer in the alpha direction. If the target is above or below then you nose up or down towards it. Otherwise if the target is on your sides, then you don't do anything but wait for the airframe to rotate.
Imagine how a fighter airplane dog fights: the pilot rolls the airplane towards the enemy then pull the nose up. RIM is very similar but different in two ways: 1) rolling is done by auto-rotation and not initiated by any control surfaces, 2) RIM can handle negative alpha and negative G much better than an airplane so it has two chances to maneuver towards the target, per revolution, instead of just one.
In practice the control low is automatically solved the closed-loop nature of the homing system. From the missile's point of view it nothing but a target that periodically moves up and down. Simply put, auto-rotating airframe turns a 1D homing system into a 2D one.


text:
Is there good information available on maximum operating altitudes whilst maintaining a sea level cabin?
I am in the process of reviewing the aircraft we regularly commission for air ambulance transfers. Although the majority of transfers are undertaken with a standard cabin altitude ( 4 - 8,000 ft) some are conducted with a sea level cabin (or cabin altitude restriction). The reasons for this are not really necessary for the purposes of the question, but the consequences of requesting one are that the operating maximum altitude of the aircraft is reduced (to maintain a safe cabin wall differential pressure) and the range / endurance of the aircraft is reduced.
Although this has low operational impact (and would normally not affect our choice of providers) I wondered if anyone had stumbled upon a list of airframes and the corresponding max altitudes that they would be limited to while maintaining a sea level cabin environment.
Any help would be much appreciated.

The definitive answer would come from the manufacturer, but you can calculate some estimates using the barometric formula in combination with known pressure differential limits for specific aircraft. Below are some examples for several aircraft types that are used as air ambulances.

Type
psid
Maximum Sea Level Altitude

King Air 90 (A90)
4.6 psi
10,000 ft

King Air 100
4.7 psi
10,200 ft

King Air 90 (C90)
5.0 psi
11,000 ft

Pilatus PC12
5.8 psi
13,200 ft

King Air 200/300
6.5 psi
15,200 ft

Learjet 35
9.4 psi
25,500 ft

Some of the psid values that I used came from the AOPA website and BeechTalk Forums.
If you are interested in how I calculated the estimates for these planes so that you can calculate estimates for other aircraft, I have provided an explanation in the next section.
Some private jets can go even higher, for example the Bombardier Global Express has a maximum pressure differential of 10.3 psi which allows a cabin pressure of 5,680 feet at its ceiling of 51,000 feet (Wikipedia). So it should be able to maintain sea level cabin pressure up to 30,000 feet. The leader currently seems to be the Syberjet SJ30i which has a maximum pressure differential of 12 psi which the company website  says allows a sea level cabin pressure up to 41,000 feet. And an honorable mention goes to Concorde which had a differential of 10.7 psi which allowed a cabin pressure of 6,000 ft at its service ceiling of 60,000 ft.
Calculation for Other Aircraft
The above estimates were calculated using the barometric formula:

Using a simplified version of the barometric formula along with a few additional steps you can calculate what should be a pretty good estimate of sea level cabin pressure maximum altitude for any aircraft for which you know the maximum psid (psi differential). By simplified version I mean with all of the several variables already filled in.
Step 1
The first step is to subtract the known maximum psid for a particular aircraft from standard sea level pressure to determine the psi at the maximum sea level altitude. For example if calculating for the King Air 100 you would subtract 4.7 psi from 14.7 psi, which tells us that the air pressure at the maximum altitude for sea level cabin pressure is 10 psi.
Step 2
So now we just need to determine what altitude has a pressure of 10 psi. There are tables online that have this information, but for more accuracy we can run the barometric formula for different altitudes until we find one that has the exact psi that we are looking for.
The simplified equation with all variables except altitude filled in looks like this:
14.6953 ((1 – (0.000006875586  h))5.25588 )
h is the altitude in feet that you are calculating the air pressure for. In this case your first guess might be 15,000 feet. You can use a scientific calculator, including the built-in Windows calculator (switched to scientific mode) and do the calculation using the following keystrokes:
MC 15000 X 0.000006875586 M+ 1 – MR = xy 5.25588 X 14.6953
which gives a result of 8.3 psi.
Even easier is to enter the formula into Excel. A1 represents the cell where you enter altitude h.
=14.6953*POWER(1-(0.000006875586*A1),5.25588)

We are looking for an altitude that has a psi of 10.0 and 8.3 is too low. So next we might try 10,000 which is 10.1 psi, then after a couple more tries arrive at 10,200 which is 10.0 psi.
Once you have done this a couple of times it goes pretty quickly, especially if using a spreadsheet. You might even use an online table to provide the first guess, and then fine tune it using the formula. Of course in the end these are just estimates, it’s up to you how accurate the estimate needs to be.
I am making an assumption that the maximum psid for flying at high altitudes can be used to establish sea level cabin pressure limits at lower altitudes. Someone may point out that the Boeing 737, which has a maximum psid of 8.3 at its service ceiling, has lower psid’s for lower altitudes. But that is for maximizing the lifetime cycles of the airframe by using a higher cabin altitude at lower altitudes than the 8.3 psi limit would otherwise allow (b737.org.uk). For example at 26,000 feet the 8.3 psi limit would allow a cabin altitude of 2,300 feet. But the maximum psid for the 737 under 28,000 feet is 7.45 psi, which requires a cabin altitude of 4,000 feet, not from a safety standpoint, but economic in terms of airframe life.
As further evidence of this, the Boeing 737 has an estimated lifespan of 75,000 cycles. The BBJ (Boeing Business Jet) however is certified with a psid of 9.0, similar to smaller business jets. This is apparently not because of any structural modifications to the BBJ compared to the airline version of the 737, but simply because they expect that in VIP use the lifetime cycles will be lower. The result of the higher psid limit is that it reduces the BBJ lifespan to 60,000 cycles instead of 75,000. This makes it pretty evident that the 8.3 psid for the 737 is not a strict structural limit, otherwise the BBJ would not have been certified to 9.0 psid.
Another example is that private jets often cruise with a cabin altitude of around 4,500 feet, however reportedly when only the pilots are onboard they often fly at a cabin altitude of 8,000 feet to increase airframe life. So it’s possible that a manufacturer might certify an air ambulance for a higher psid than the normally published limit, with the condition that the number of lifetime cycles will be reduced.
Other Altitudes
Besides calculating sea level cabin pressure maximum altitude this method can also be used to calculate other cabin altitudes that may be of interest. I read your comments in the other question about the medical reasons for the 8,000 ft cabin pressure limit for airliners which was quite interesting. As you may know there are different rules for general aviation. Even though judgment can be affected above 10,000 feet, FAA regulations for general aviation  don’t begin until 12,500 feet, and even then only if pilots spend more than 30 minutes above that altitude. Above 14,000 feet pilots must use oxygen at all times, and above 15,000 feet passengers must be provided with oxygen. It’s interesting (and useful) to calculate at what altitude a particular pressurized airplane will have a cabin altitude of for example 10,000 feet.  Using the King Air 100 again as an example,
Step 1
Determine the pressure at 10,000 feet. Using the formula this is determined to be 10.1 psi.
Step 2
Subtract the King Air 100 psid of 4.7 from 10.1, the result is 5.4 psi.
Step 3
Use the formula to determine what altitude has a psi of 5.4. The result is 25,250 ft. So that would be the estimated altitude where the King Air 100 would have a cabin altitude of 10,000 feet. Of course in actual flying pilots will be using their pressure instruments and/or tables to make this determination, but still it’s interesting to do some estimates.
Along the lines of your question, air ambulances seem to sometimes operate at higher than sea level cabin pressure, but they seem to be limited to a 5,000 foot cabin altitude or similar. Here are some estimates for our example aircraft where I have included 5,000 and 8,000 foot cabin altitudes, along with the service ceilings that I could find for these airplanes.

Type
psid
Cabin 0
Cabin 5000
Cabin 8000
Service Ceiling

King Air 90 (A90)
4.6 psi
10,000 ft
17,000 ft
21,500 ft
30,200 ft

King Air 100
4.7 psi
10,200 ft
17,500 ft
22,000 ft
24,850 ft

King Air 90 (C90)
5.0 psi
11,000 ft
18,500 ft
23,000 ft
29,800 ft

Pilatus PC12
5.8 psi
13,200 ft
21,000 ft
26,500 ft
30,000 ft

King Air 200/300
6.5 psi
15,200 ft
24,000 ft
29,500 ft
35,000 ft

Learjet 35
9.4 psi
25,500 ft
39,000 ft
52,000 ft
45,000 ft

The Equations
Optional reading in case you are interested in how the barometric formula works. I am using the equations (plural) shown in Wikipedia.
The reason that there are two equations is because at some altitude ranges the temperature does not change much with altitude, so they have a different formula for those altitude ranges. Fortunately the first altitude range is from 0 to 11,000 meters (36,089 feet), which covers a high percentage of airline flights. The second range, using a different formula, is from 11,000 to 20,000 meters (36,089 to 65,617 feet). However for the first several thousand feet of that range the calculation results are very similar. So the 0-FL36 formula is generally adequate for all commercially flown altitudes. That’s why at the beginning of this answer I only mentioned one formula.
Using 41,000 ft as an example, using the correct FL36-FL65 formula for that altitude gives a result of 2.59 psi. Using the 0-FL36 formula gives a result of 2.58 psi. You do start to see some differences however when you get up to Concorde height at 60,000 feet. Using the correct FL36-FL65 formula for that altitude gives a result of 1.04 psi. Using the 0-FL36 formula gives a result of 0.90 psi which is a 13% difference.
I will explain both formulas here, but I will only go into detail on the 0-FL36 formula, since the main points apply to both.
As mentioned before the 0-FL36 equation looks like this:

Some of these are a fixed constant, for others the Wikipedia page includes a table for the variables for the different altitude ranges. The formula works with either metric or imperial units, and the tables in Wikipedia have values for both. For this explanation I am using imperial units.
Hb – the reference altitude. For the 0-FL36 formula that would be sea level or 0 feet. For the FL36-FL65 formula the reference altitude is 11,000 meters, or 36,089 feet.
h – this is the altitude that we are calculating the pressure for. After simplifying this will be the only variable left in the formula.
Tb  – the standard temperature at the reference altitude in kelvin, for the 0-FL36 formula this is 288.15 K (59 °F)
G0 – gravitational acceleration, this is a fixed constant  at 32.17 ft/s2
M - molar mass of the atmosphere, a fixed constant  28.96 lb/lbmol
R – universal gas constant, a fixed constant 8.949×104 lbft2/(lbmol•K•s2)
Pb – the standard pressure at the reference altitude and temperature, for the 0-FL36 formula this is 29.92 inHg (inches of mercury).
Lb – temperature lapse rate due to altitude change, for the 0-FL36 formula this is 0.001981 K/ft (yes that unit works)
P – the result in inHg, which we will then convert to psi.
With the constants and variables inserted the equation now looks like this:

The result in inHg is then multiplied by 0.4912 to get psi. I won’t go through the steps of simplifying but this is the resulting simplified formula that was mentioned earlier:
14.6953 ((1 – (0.000006875586  h))5.25588 )
And as mentioned the Excel version of the formula would be:
=14.6953*POWER(1-(0.000006875586*A1),5.25588)
Running the equation for 0 altitude we get 14.7 psi as expected. Looking at some other altitudes
 5,000 ft = 12.2 psi
 6,000 ft = 11.8 psi
 8,000 ft = 10.9 psi
10,000 ft = 10.1 psi
20,000 ft =  6.8 psi
35,000 ft =  3.5 psi

For altitudes between 36,089 and 65,617 feet there is an exponential function used which assumes that the temperature does not drop with altitude:

And there are different values for some of the variables.
Here is the simplified version of the FL36 - FL65 formula:
3.28281 e(-0.0000480634 h) + 1.73456)
Using a scientific calculator, using 41,000 feet for example this would be:
MC 41000 X 0.0000480634 +/– + 1.73456 = M+ e xy MR = X 3.28281
with a result of 2.6 psi.
The Excel version is:
=3.28281*EXP((-0.0000480634*A1+1.73456))


text:
Low and High G snapshot
I'm reading Fighter Combat: Tactics and Maneuvering by Robert L. Shaw and ran into this passage which I'm having trouble understanding. I was thinking maybe someone with experience would understand better what it meant, specifically the "roll back in the opposite direction to get into the target's plane of turn". I'm thinking i might not be imagining the way the author intented me to do so since I can't understand what that means. Thanks in advance and sorry if I made a mistake with the posting of the question, is the first time I use this site

First you have to understand the issues with a high (more than 60deg) aspect angle gunshot. At these high aspect angles, if the target is at high speed, his Line of Sight (LOS) rate across your nose can easily exceed your maximum turn rate. This is what makes it a "snap" shot. You cannot "Track" the target. In addition, because of the high LOS rate, in order to "join up" to a position where the target will fly through the gun bore line within effective gun range (1000-2000 feet), it will require initially pulling so much lead that if your flight path vector is in the same 2D plane as the target is turning in, his aircraft will be blocked out by your radome, and you will lose sight.

Therefore, you have the two options described.

Option 1. You can initially pull more lead (almost to the collision course), and than turn with substantially less turn rate and G-load, while you close into gun range. If you are turning in plane, this would entail losing sight under the radome. You could do this and then continuously roll out of the turn momentarily to reacquire the tally, over and over, as you close. But generally, this is problematic, and the more recommended option is to perform the initial manuever by holding the target barely to the side of or on the edge of the radome, (out of plane), until you have closed sufficiently that the target LOS rate starts to increase - to where he begins drifting forwards towards the nose. Then, slide sideways slightly to bring the target back "in plane" so that his LOS motion will bring his aircraft through the gun bore line and the bullet stream. This option will generally result in a much higher LOS rate gunshot, and expose him to fewer bullets. But because it will be done with higher angle off (heading crossing angle), it allows for a safer overshoot and separation maneuver.
Option 2. Or, you can initially pull less lead, in plane, keeping the target just barely visible on the edge of the random, in order to maintain Tally-Ho, and then gradually increase the back pressure, turn rate and G-load, to keep the target on the edge of the radome, until the targets LOS rate, (and your turn rate to track it) has increased to just below your maximum. This generally occurs just outside the max effective gun range, (around 2500-3000 ft). Then pull harder to try to match the target's LOS rate just before you turn on the gun. You will not be able to match the targets LOS rate, but then, (if you are still in plane), he will fly through the gun bore line and the bullet stream. Using this option, you will arrive at this position (with the target flying through the bullet stream), with less angle off and Target LOS rate, and will potentially expose the targets to the bullet stream for a longer period of time, (more bullets), than using the other option.  This option (because you will have less angle off (Heading crossing angle), has the disadvantage of making the overshoot and separation maneuver more difficult, and, against a more maneuverable adversary, more dangerous.

Thank you for the clear explanation


text:
Could a jet engine theoretically ingest not only oxidizer but fuel (on planets like jupiter)
Jet engines work by ingesting oxygen from the air and combusting it with an onboard fuel source like kerosene. But, on planets like Jupiter, with an abundance of hydrogen, would it be possible to remove the onboard fuel source and ingest both oxygen and hydrogen, carrying no fuel at all? This is assuming that the plane would be able to withstand very high pressures, and if Jupiter doesn't have the right amount of oxygen, any other planets that have the right atmosphere to support it.

"any other planets that have the right atmosphere": There are none in the solar system. Earth with its load of O2 is really an exception, thanks to photosynthesis (GOE). There is no other known place where a significant quantity of 02 can be found.

Jupiter has hydrogen, but not oxygen. So you would need to bring your own oxygen, and essentially have just replaced what you use as "fuel."
If the atmosphere was, by itself, flammable then a jet engine should work. For a brief period of time before you set the atmosphere on fire.
Atmospheres don't tend to be flammable, for what are pretty obvious reasons if you think about it.

Wouldn't an atmosphere of pure flammable gas with no oxygen be safe for a jet engine that burned atmospheric gas and added oxygen in the way Earth jets add fuel, since after combustion in the jet, there would be no oxygen to burn the atmosphere further?

@Someone I think that's the point of the first paragraph here. An atmosphere with fuel but no oxygen (Jupiter's) can exist, and your aircraft would have to provide its own O2. An atmosphere with oxygen but no fuel (i.e. Earth's) can exist, and the aircraft brings its fuel. An atmosphere with both fuel and oxygen can exist until the first spark, and then... well, you'll burn up the fuel until it or the oxygen runs out, and then you're back to one of the first two cases.

Not sure if it would matter as the ratio would be the same when compressed, but maybe there could be a mix of fuel and oxygen such that at normal pressure (whatever that is on planet X) it does not normally burn, but that when compressed in a jet engine it burns well.

@manassehkatz-Moving2Codidact Maybe. The possibility that came to mind for me is a very very low density atmosphere wouldn't catch fire under any circumstances simply because there is too much space between the molecules. Perhaps a ramjet could make use of such an atmosphere.

In any case even it such an atmosphere could exist (which I'm not sure about), I'm pretty certain that no such atmosphere does exist in our solar system.

Maybe at a sufficiently high altitude, you could have such a thin atmosphere that a flame couldn't propagate, but at lower altitude you'd have higher pressure, greater density, and the same first-spark-ignites-the-atmosphere situation. This may be approaching a question better on Worldbuilding.SE than here.

I very much doubt such an atmosphere could exist even at low density--even if it can't support flame you will get reactions sometimes when molecules bump into each other.  Thus it would not be stable.

@LorenPechtel: …and yet the atmospheric gas mix doesn't need to be long term stable — it just needs to react slowly enough that other processes (biology, volcanism, photochemistry, etc.) can replenish it. After all, the Earth's atmosphere isn't thermodynamically stable either: it's mostly composed of dinitrogen and dioxygen, which react very slowly to form various nitrogen oxides. But the timescale for this reaction is longer than the biological cycle time for these gases. (Of course that's an endothermic reaction, so you can't really use it to drive an engine even if you catalyzed it.)

it is the Earth who has both hydrogen and oxygen in the atmosphere, the problem is that they are separated by tens of km elevation. The thermosphere and the exosphere are hydrogen-dominated, near the surface we have the nitrogen/oxygen mixture we are used to. And while I am not sure what the transition layer looks like, I am confident that no part of our atmosphere is able to burn by itself, no matter how much compressed. It is the photochemistry that will delete anything like this on a short timescale even if no sparks are available.

p.s. Jupiter is quite aviation-unfriendly anyway. Too much gravity, too little lift vs drag (hydrogen), too much turbulence, way too much clouds and no subsance good for filling a baloon that has to be lighter than the surrounding gas.

Feels like it's time to take out the fusion ramjet and build a plane that runs in Jupiter's upper atmosphere.

the atmosphere was, by itself, flammable - you see the problem there, right?

To elaborate on the other answer, you don't actually need any fuel or oxidizer to run a jet engine. Jet engines are just a type of open-cycle heat engine where the atmosphere is the working fluid and thrust happens by expelling it as reaction mass. Usually this is implemented as a form of internal combustion engine— Fuel burns inside, that provides the heat, the exhaust shoots out the back, and you go forwards fastly, very speed.
But you can completely remove the fuel-burning step, and replace it with any equivalent source of heat, and it would still be a fully functional jet engine. Nuclear reactors have been tried— No oxygen-burning fuel at all, just a bundle of uranium that the air goes around. Or if you cut a transparent window out of the side of the combustion chamber, and you shone a strong enough laser into it, that could work too.
On Jupiter, you would indeed need to bring your own oxygen/oxidizer, and ingest hydrogen as fuel.

Or if we're talking science fiction, "burn" the hydrogen with a fusion reaction, no oxygen needed.  Only a tiny fraction of the hydrogen has to fuse to heat the rest of it a lot.  (But you'd need some kind of cold fusion shenanigans or else that airflow would cool the plasma too much...)

@PeterCordes Hm, indeed. Wouldn't strictly need to be cold fusion shenanigans either; You could localize the fusion to small volumes for very brief durations (E.G. pulsed Z-pinch powered by the turbine), as long as you're not overly insistent on continuous self-sustaining ram-compression-burning.

Yep. Even a battery attached to resistive heating elements would theoretically work. Though finding batteries with high enough capacity that can handle that much power draw will be a challenge (and the battery weight will be far more than carrying a bunch of kerosene or oxygen.)

that's clever, @PeterCordes !!

@PeterCordes Sadly ordinary hydrogen (protium) cannot be fused with protium except by relying on an incredibly rare weak reaction that converts one of the protons into a neutron, which makes a natural hydrogen fusion reactir extremely impractical. Well, I say "sadly" but it keeps our sun from burning all of its fuel extremely fast and going out in a blaze of glory, so perhaps it has its positives as well. ;)


text:
Does speed affect the pressure and lift distribution at the wingtip?
So I was wondering if speed affects the pressure distribution at the wingtip. The second comment on this answer says (if I interpret it correctly) that speed doesn't change the pressure distribution at the wingtip, but why is that?
My thinking is that if you go faster, the air outside of the tip wouldn't have as much time to react to the low pressure zone at the tip because it's going faster. Kind of like in orbital mechanics, if you go faster by a planet your orbital line doesn't change as much because you're going faster (less time under acceleration). If the air doesn't have as much time to react, then wouldn't less air be bent towards the wingtip, affecting the pressure (and lift) distribution?
Thanks!

Small pressure changes travel at the speed of sound. In sufficiently subsonic flow the pressure distribution at different speeds is self-similar (same shape, different magnitude). Only when tip flows become transsonic will you see a marked change in pressure distribution. This change will become fundamental once the speed component orthogonal to the wing's leading edge becomes supersonic.
See it this way: The faster-moving wingtip will produce pressure variations of greater magnitude, so the air will experience more force over a shorter time. Both effects cancel each other and the flow pattern stays the same. Your gravity analogy would need increasing gravity with higher orbital speed to be suitable here.

ah that makes sense, thanks a lot. One small question : when the wing reaches transsonic speeds, what would make the pressure distribution change? (I'm assuming the transsonic speeds you're talking about is around mach 0.6, because of other answer I read)

@Wyatt Yes, the transsonic regime starts strictly at Mach 0.3, but the effects only become noticeable at around Mach 0.5 to 0.7, depending on local pressure. Please read this answer to get an idea what changes. Essentially, disturbances will reach further sideways and once local speed exceeds the speed of sound, it will accelerate further where in subsonic flow it would decelerate. Local shocks will do that deceleration eventually.

ah, okay thanks. I read the answer you linked and it clarified some things. One last question though : What makes disturbances reach further sideways with higher speeds? My thinking was that lower density air (from higher speed) would prevent disturbances from reaching as far. (I did read the tube analogy in the linked answer)

@Wyatt Since density drops at higher subsonic speeds, more tubes need to be involved at making way for the object. Each tube contracts less, so more of them need to accelerate their flow for the needed contraction.

ah okay that clarifies things, thanks. I have another question posted here (related to this question) and would greatly appreciate your insights on that as well, if you don't mind. Thanks in advance!


text:
TSA clearance for pilot training
Hi everyone I have my tsa clearance for instrument in between my instrument I am starting my commercial training do I need to get a separate tsa clearance for that ?

This sounds like a question for [Aviation.SE].

I’m voting to close this question because it is not about travel as defined in the help center.

I was told that here a lot of tsa questions are answered

You can ask in the aviation stackexchange. TSA, in context of travel is definitely answered here.

TSA flight training clearances are not specific to a rating or endorsement. Unless your clearance specifically states that you are only allowed to use it to get an FAA instrument rating, you can use it for any flight training.
That said, take the document to the flight training school where you want to get your commercial ticket and ask them this question. They are the ones who will have to accept the document so they are the only ones who can tell you if it's good for that, or not.


text:
ATC gives wrong instruction, and you follow it
Example I saw recently from a pilot I know:
He was inside a VIP TFR and had filed a flight plan, picked up a code , and was talking to atc.   Inside the TFR he reached his destination and atc told him  to change to advisory and squawk VFR.  Later the FBO got a call saying the jets were nearly scrambled due to this mistake.  Who’s at fault here ?

I’m assuming the squawk VFR while inside the TFR. Considering he received a call that the jets were almost scrambled because of him

It depends on the wording of the TFR, but most likely they both made a mistake. Most VIP TFRs contain language similar to:

ALL ACFT MUST BE ON AN ACT IFR OR FILED VFR FLT PLAN WITH A DISCRETE CODE ASSIGNED BY AN AIR TFC CTL (ATC) FAC. ACFT MUST BE SQUAWKING THE DISCRETE CODE PRIOR TO DEP AND AT ALL TIMES WHILE IN THE TFR AND MUST REMAIN IN TWO-WAY RADIO COM WITH ATC.

Or for the smaller TFRs:

ACFT MUST BE SQUAWKING AN ATC DISCRETE BEACON CODE AT ALL TIMES WHILE IN THE TFR AND MUST REMAIN IN TWO - WAY RADIO COM WITH ATC.

In either case, it doesn't have any language about "Unless authorized by ATC." You need to have a discrete squawk code at all times in the TFR. ATC should not have told the pilot to squawk VFR, and the pilot should have refused the code change.

Moreover, as I read 14 CFR 91.141 - The pilot's the one most likely to end up in the worst trouble, to the extent any trouble occurs. The regulation says "No person may operate an aircraft...contrary to the restrictions..." ATC - while not helpful here - wasn't operating the aircraft.

From the VIP TFR: "EXC AS SPECIFIED BLW AND/OR UNLESS AUTH BY ATC IN CONSULTATION WITH THE AIR TFC SECURITY COORDINATOR VIA THE DOMESTIC EVENTS NETWORK (DEN) [blah blah blah, here are the exceptions]." So ATC can in fact authorize other non-specified exceptions. Whether they did the required coordination is not clear—probably they did not, given the FBO's comment—but how is the pilot to know that? The pilot fully complied with the TFR.

@randomhead That language is unusual in VIP TFRs, but it can appear and is why I was careful to say "most" and not "all." Of the 7 currently published VIP TFRs that require you to squawk a discrete code, only one has "Unless auth by atc" anywhere in it. I agree that the pilot technically complied with the TFR if that language appeared in the specific TFR in question, but a quick "Confirm N12345 squawk VFR in the TFR" would certainly be warranted.

Fair points, touché. I do wonder if the difference is between POTUS VIP TFRs and VPOTUS and/or candidate VIP TFRs, because I'm only seeing one active POTUS TFR at the moment. (And then there's the George W. Bush TFR near Dallas Love, which also says "unless otherwise authorized by ATC.")

A careful analysis of what you wrote suggests no one did anything wrong. The military jets were likely not scrambled because someone called ATC and found out the pilot did exactly what he was told to do.

Why couldn't it be an ATC mistake? Say civilian ATC was told to keep everybody under radio contact and positive control the controller or facility messed up.

That's a different question, for which you have not included enough information to form an opinion.


text:
Why aren't there rotorcraft with sailcloth rotors?
I was wondering why no one has built rotorcraft, e.g a gyrocopter with rotors made of sailcloth fabric instead of aluminium or composite, because I thought that such a rotor would have to spin at a slower rpm to generate equal lift force or generate more lift, so the craft would need less engine power to lift off. I thought of this concept after watching a human powered helicopter lift off which had large sailcloth rotors that spinned slowly.

https://aviation.stackexchange.com/questions/99522/could-floppy-rotor-blades-be-usable

Got a link to the human powered helicopter?  And why might a sailcloth rotor generate more or equal lift at slower rpm?

Here:  https://www.youtube.com/watch?v=syJq10EQkog&pp=ygUYaHVtYW4gcG93ZXJlZCBoZWxpY29wdGVy


text:
Find the approximate base of the cumulus clouds with Lapse Rate
Okay, so I know how to calculate the base of cumulus clouds, however, this is the first time I'm seeing lapse rate in the question:
Normally I would do ((23 - 15) / 2.5) * 1000 which would give me 3200, then add the elevation, which was 3700. I figured 4,000 was the "close" answer due to lapse rate, but I was wrong.

Any help would be appreciated

Way out of my stovepipe, but don't they tell you to use 2 for the lapse rate and you used 2.5? Which would make the answer 4500.

Wow, now I feel silly. That makes perfect sense now. That was the issue.

OK, I'll post it as an answer.

The lapse rate given in the question is 2 but you used 2.5.
Substituting the proper lapse rate gives an answer of 4500.